Problem: $h(x) = 4x+3$ $f(t) = 6t^{2}+4t-3(h(t))$ $ f(h(0)) = {?} $
Answer: First, let's solve for the value of the inner function, $h(0)$ . Then we'll know what to plug into the outer function. $h(0) = (4)(0)+3$ $h(0) = 3$ Now we know that $h(0) = 3$ . Let's solve for $f(h(0))$ , which is $f(3)$ $f(3) = 6(3^{2})+(4)(3)-3(h(3))$ To solve for the value of $f$ , we need to solve for the value of $h(3)$ $h(3) = (4)(3)+3$ $h(3) = 15$ That means $f(3) = 6(3^{2})+(4)(3)+(-3)(15)$ $f(3) = 21$